Maths M.C.Q (1 Marks)

$ = 2\,[1 + {\,^9}{C_2}{(3\sqrt 2 x)^2} + {\,^9}{C_4}{(3\sqrt 2 x)^4} + {\,^9}{C_6}{(3\sqrt 2 x)^6} + {\,^9}{C_8}{(3\sqrt 2 x)^8}]$

The number of non-zero terms is $5.$

${(101)^{100}} - 1 = 100.100\left[ {1 + \frac{{100.99}}{{1.2}} + \frac{{100.99.98}}{{1.2.3}}.100 + ....} \right]$

From above it is clear that,

${(101)^{100}} - 1$ is divisible by $(100)^2$ $= 10000$

Now, ${7^{300}} = {({7^2})^{150}} = {(50 - 1)^{150}}$

= $^{150}{C_0}{(50)^{150}}{( - 1)^0} + {\,^{150}}{C_1}{(50)^{149}}{( - 1)^1} + ...... + {\,^{150}}{C_{150}}{(50)^0}{( - 1)^{150}}$

Thus the last digits of ${7^{300}}$ are $^{150}{C_{150}}.1.1$ $i.e., 1$.

$^n{C_6}{x^{n - 6}}{a^6} + .......]$

Here, $n = 6,x = \sqrt 2 ,a = 1$; $^6{C_2} = 15,{\,^6}{C_4} = 15,{\,^6}{C_6} = 1$

$\therefore \,\,{(\sqrt 2 + 1)^6}{(\sqrt 2 - 1)^6} = 2[{(\sqrt 2 )^6} + 15.{(\sqrt 2 )^4}.1$
$ + 15{(\sqrt 2 )^2}.1 + 1.1]$

$ = 2[8 + 15 \times 4 + 15 \times 2 + 1] = 198$

Hence ${T_6} = {\,^{10}}{C_5}{(2{x^2})^5}{\left( { - \frac{1}{{3{x^2}}}} \right)^5}$

$ = - \frac{{10\,!}}{{5\,!\,5\,!}}32 \times \frac{1}{{243}} = - \frac{{896}}{{27}}$

But according to the condition,

$\frac{{ - \,n(n - 1) \times 3 \times 2 \times 1 \times 8}}{{n(n - 1)(n - 2) \times 2 \times 1 \times 4}} = \frac{1}{2}$

$\Rightarrow n = - 10$

$ \Rightarrow $ $20 - r + 1 = r + 3$

$\Rightarrow r = 9$.

$ = \frac{{n!}}{{(n - r + 1)!(r - 1)!}}{a^{n - r + 1}}{(2x)^{r - 1}}$

$ = \frac{{n(n - 1).....(n - r + 2)}}{{(r - 1)!}}{a^{n - r + 1}}{(2x)^{r - 1}}$

$\frac{n}{3} - 4 = - 1$

$ \Rightarrow $ $n = 9.$

But $^{15}{C_{2r + 2}} = {\,^{15}}{C_{15 - (2r + 2)}} = {\,^{15}}{C_{13 - 2r}}$

==> $^{15}{C_{13 - 2r}} = \,{\,^{15}}{C_{r - 2}}$

$\Rightarrow r = 5$.

==> $67 - 7r = 4 $

$\Rightarrow r = 9$.

By hypothesis, $\frac{{m(m - 1)}}{2}{x^2} = - \frac{1}{8}{x^2}$

==> $4{m^2} - 4m = - 1$

==> ${(2m - 1)^2} = 0$

$\Rightarrow m = \frac{1}{2}$.

${T_2} = {}^n{C_1}ax = 6x$ …..$(ii)$

${T_3} = {}^n{C_2}{(ax)^2} = 16{x^2}$ …..$(iii)$

From $(ii)$, $\frac{{n!}}{{\left( {n - 1} \right)\,!}}a = 6$ 

==> $n\,a\, = \,6$ …..$(iv)$

From $(iii)$, $\frac{{n(n - 1)}}{2}\,{a^2} = 16$ ….$.(v)$

Only $(c)$ is satisfying equation $(iv)$ and $(v)$.

By the given condition $2\,{.^{14}}{C_r} = {\,^{14}}{C_{r - 1}} + {\,^{14}}{C_{r + 1}}$ …..$(i)$

==> $2\,.\frac{{14!}}{{r!\,\,(14 - r)!}}\, = \,\frac{{14!}}{{(r - 1)!\,\,(15 - r)!}} + \frac{{14!}}{{(r + 1)!\,(13 - r)!}}$

==> $\frac{2}{{r.(r - 1)!.(14 - r).(13 - r)!}}$=$\frac{1}{{(r - 1)!.(15 - r).(14 - r).(13 - r)!}}$

+$\frac{1}{{(r + 1)\,r(r - 1)\,!\,(13 - r)\,!}}$

==> $\frac{2}{{r(14 - r)}} = \frac{1}{{(15 - r)(14 - r)}} + \frac{1}{{(r + 1)r}}$

==> $\frac{1}{{r(14 - r)}} - \frac{1}{{(15 - r)(14 - r)}} = \frac{1}{{(r + 1)r}} - \frac{1}{{r(14 - r)}}$

==> $\frac{{(15 - r) - r}}{{r(15 - r)(14 - r)}} = \frac{{(14 - r) - (r + 1)}}{{(r + 1)r(14 - r)}}$

==> $\frac{{15 - 2r}}{{15 - r}} = \frac{{13 - 2r}}{{r + 1}}$

==> $15r + 15 - 2{r^2} - 2r = 195 - 30r - 13r + 2{r^2}$

==> $4{r^2} - 56r + 180 = 0\,\,$

$\Rightarrow \,{r^2} - 14r + 45 = 0$

==> $(r - 5)(r - 9) = 0 $

$\Rightarrow r = 5,\,9$

But $5$ is not given.

Hence $r = 9.$ 

$Trick :$ Put the value of $r$ from options in equation $(i)$, only $(d)$ satisfy it.

$= {\,^5}{C_r}{a^r}{x^{10 - 3r}}$

Here, exponent of $x$ is $10 - 3r = 1$

$\Rightarrow r = 3$$\therefore $ ${T_{2 + 1}} = {\,^5}{C_3}{a^3}x = 10{a^3}.x$

Hence coefficient of $x$ is $10{a^3}$.

$ \Rightarrow - 12 + r + r = - 10 $

$\Rightarrow r = 1$ 

Then coefficient of ${x^{ - 10}}$ is $^{12}{C_1}{(a)^{11}}{(b)^1} = 12{a^{11}}b$.

$ = \frac{{^{2n}{C_n}}}{{^{(2n - 1)}{C_n}}} = \frac{{(2n)!}}{{n!\,n!}} \times \frac{{(n - 1)!\,n!}}{{(2n - 1)!}}$

$ = \frac{{(2n)(2n - 1)!(n - 1)!}}{{n(n - 1)!\,\,(2n - 1)!}} = \frac{{2n}}{n} = 2:1$

==> $\frac{A}{B} = \frac{2}{1}$

==> $A = 2B$.

According to the condition, $2\,{\,^n}{C_5} = {\,^n}{C_4} + {\,^n}{C_6}$

$ \Rightarrow \,\,2\left[ {\frac{{n!}}{{(n - 5)!5!}}} \right] = \left[ {\frac{{n!}}{{(n - 4)\,!\,4\,!}} + \frac{{n!}}{{(n - 6)\,!\,6\,!}}} \right]$

$ \Rightarrow \,\,2\left[ {\frac{1}{{(n - 5)\,5}}} \right] = \left[ {\frac{1}{{(n - 4)(n - 5)}} + \frac{1}{{6 \times 5}}} \right]$

After solving, we get $n=7$ or $14$.

= ${}^{2n}{C_r}{x^{2n - 3r}},$

This contains $x^m$, if $2n -3r = m\ i.e$. if $r = \frac{{2n - m}}{3}$  

Coefficient of $x^m$ $ = {}^{2n}{C_r},$ $r = \frac{{2n - m}}{3}$ = $\frac{{2n!}}{{(2n - r)!r!}}$

=$ \frac{{2n!}}{{\left( {2n - \frac{{2n - m}}{3}} \right)\,\,!\left( {\frac{{2n - m}}{3}} \right)\,\,!}}$

= $\frac{{2n!}}{{\left( {\frac{{4n + m}}{3}} \right)\,\,!\,\,\left( {\frac{{2n - m}}{3}} \right)\,\,!}}$.

==> ${2.^n}{C_2} = {\,^n}{C_1} + {\,^n}{C_3}$ 

==> $\frac{{2n!}}{{2\,!\left( {n - 2} \right)\,!}} $

= $\frac{{n!}}{{(n - 1)!}} + \frac{{n!}}{{3!\,\left( {n - 3} \right)\,!}}$ On solving, ${n^2} - 9n + 14 = 0\,\,$

$\Rightarrow \,\,{n^2} - 9n = - 14$.

$ = {(1 + x)^{21}}\left[ {\frac{{{{(1 + x)}^{10}} - 1}}{{(1 + x) - 1}}} \right]$

= $\frac{1}{x}[{(1 + x)^{31}} - {(1 + x)^{21}}]$  

$\therefore$ Coefficient of $x^5$ in the given expression 

= Coefficient of $x^5$ in $\left\{ {\frac{1}{x}[{{(1 + x)}^{31}} - {{(1 + x)}^{21}}]} \right\}$ 

= Coefficient of $x^6$ in $[{(1 + x)^{31}} - {(1 + x)^{21}}]$ = ${}^{31}{C_6} - {}^{21}{C_6}$.

Coefficient of $T_2, T_3, T_4$ are in $A.P$. 

==> $2.{}^{2n}{C_2} = {}^{2n}{C_1} + {}^{2n}{C_3}$ 

==> $2\frac{{2n!}}{{2\,!\,(2n - 2)\,!}} = \frac{{2n!}}{{(2n - 1)\,!}} + \frac{{2n!}}{{3\,!\,(2n - 3)\,!}}$

==> $\frac{{2\,.\,2n(2n - 1)}}{2} = 2n + \frac{{\,2n(2n - 1)(2n - 2)}}{6}$ 

==> $n(2n - 1) = n + \frac{{(n)(2n - 1)(2n - 2)}}{6}$ 

==> $6(2{n^2} - n) = 6n + 4{n^3} - 6{n^2} + 2n$ 

==> $6n(2n - 1) = 2n(2{n^2} - 3n + 4)$ 

==> $6n - 3 = 2{n^2} - 3n + 4$ 

==> $0 = 2{n^2} - 9n + 7$

==> $2{n^2} - 9n + 7 = 0$.

$\therefore $ Coefficient of ${x^r}$= ${}^9{C_r}{3^{9 - r}}{a^r}$

Hence, coefficient of ${x^2} = {}^9{C_2}{3^{9 - 2}}{a^2}$ and coefficient of ${x^{\rm{3}}}$ 

= ${}^9{C_3}{3^{9 - 3}}{a^3}$

So, we must have ${}^9{C_2}{3^7}{a^2} = {}^9{C_3}{3^6}{a^3}$ 

==> $\frac{{9.8}}{{1.2}}.3 = \frac{{9.8.7}}{{1.2.3}}.a\,\,\, $

$\Rightarrow \,\,a = \frac{9}{7}.$

Here, $N = 2n +1$==> $\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{2n + 2 - r}}{r}.x$

$\therefore $ ${T_{r + 1}} \ge {T_r}$

$ \Rightarrow $ $2n + 2 - r \ge r$

$ \Rightarrow $ $2n + 2 \ge 2r$ ==> $r \le n + 1$

$\therefore \,\,\,\,\,r = n$${T_{r + 1}} = {T_{n + 1}} = {\,^{2n + 1}}{C_{n + 1}}$

$ = \frac{{(2n + 1)\,!}}{{(n + 1)!\,n!}}$.

Now first term $^{1024}{C_0}{\left( {{5^{1/2}}} \right)^{1024}} = {5^{512}}$ (integer)

And after $8$ terms, the $9^{th}$ term ${ = ^{\,\,\,1024}}{C_8}{({5^{1/2}})^{1016}}{({7^{1/8}})^8}$ = an integer

Again, $17^{th}$ term =$^{1024}{C_{16}}{({5^{1/2}})^{1008}}{({7^{1/8}})^{16}}$
= An integer.

Continuing like this, we get an $A.P.$ $1, 9, 17, ...., 1025,$

because $1025^{th}$ term = the last term in the expansion

$ = {\,^{1024}}{C_{1024}}{\left( {{7^{1/8}}} \right)^{1024}} = {7^{128}}$(an integer)

If $n$ is the number of terms of above $A.P$. we have

$1025 = {T_n} = 1 + (n - 1)8\,\,\, \Rightarrow n = 129$.

Obviously, $r$ should be a multiple of $6.$

Total numbers = $\frac{{642}}{6} = 107$; But first term for $r = 0$ is also integral. Hence total terms are $107 + 1 = 108$.

Put . $^5{C_2} = 10\,\,\,\,\,[{\log _{10}}10 = 1]$

If $x = 10$, then ${10^3}{.10^{2.1}} = {10^5}$ is satisfied.

Hence $ x =10.$

Since $(n+1)^{th}$ and $(n+2)^{th}$ terms are middle terms in the expansion of $(1+x)^{2n+1}$, therefore $q = {\,^{2n + 1}}{C_n}$ and $r = {\,^{2n + 1}}{C_{n + 1}}$ But $^{2n + 1}{C_n} + {\,^{2n + 1}}{C_{n + 1}} = {\,^{2n + 2}}{C_{n + 1}}$

$\therefore \,\,\,q + r = p$

For coefficient of $x$, $10 - 2r - r = 1\,\,\,\, \Rightarrow r = 3$

Hence, ${T_{3 + 1}} = {}^5{C_3}{({x^2})^{5 - 3}}{\left( {\frac{k}{x}} \right)^3}$

According to question, $10\,{k^3} = 270\,\, \Rightarrow \,\,k = 3$.

Hence the middle term is ${10^{th}}$.

So the term independent of $y$

= $^9{C_3}{({y^{ - 1/6}})^6}{( - {y^{1/3}})^3} = - 84$.

$\frac{{{T_2}}}{{{T_3}}} = \frac{{^n{C_1}{a^{n - 1}}b}}{{^n{C_2}{a^{n - 2}}{b^2}}}$ ......$(ii)$
$(i)$ = $(ii)$ ==> $\frac{{2n}}{{n(n - 1)}} = \frac{{6(n + 3)(n + 2)}}{{2(n + 3)(n + 2)(n + 1)}}$

==> $2(n + 1) = 3(n - 1) \Rightarrow n = 5$.

${S_1} = 1,\,\,{S_2} = 2 + 2 = 4$

Now by alternate $(c)$, put $n = 1,\,\,2$

 ${S_1} = {1.2^0} = 1,{S_2} = {2.2^1} = 4$

=$\frac{1}{{n + 1}}\left[ {(n + 1) + \frac{{(n + 1)n(n - 1)}}{{3!}} + \frac{{(n + 1)n(n - 1)(n - 2)(n - 3)}}{{5!}} + ....} \right]$

Put $n + 1$=N = $\frac{1}{N}\left[ {N + \frac{{N(N - 1)(N - 2)}}{{3!}} + \frac{{N(N - 1)\,(N - 2)(N - 3)(N - 4)}}{{5!}} + ....} \right]$

$ = \frac{1}{N}\left\{ {{\,^N}{C_1} + {\,^N}{C_3} + {\,^N}{C_5} + ....} \right\}$

$ = \frac{1}{N}\left\{ {{2^{N - 1}}} \right\} = \frac{{{2^n}}}{{n + 1}}$   $\{ N = n + 1\} $

Trick : Put $n=1$, then ${S_1} = \frac{{^1{C_0}}}{1} = \frac{1}{1} = 1$

At $n=2$, ${S_2} = \frac{{^2{C_0}}}{1} + \frac{{^2{C_2}}}{3} = 1 + \frac{1}{3} = \frac{4}{3}$

Also $(c)$ $ \Rightarrow \,\,\,{S_1} = 1,{S_2} = \frac{4}{3}$

So given term can be written as

$\frac{1}{N}\left\{ {{\,^N}{C_1} + {\,^N}{C_2} + {\,^N}{C_3} + ....} \right\}$

= $\frac{1}{N}\left\{ {{2^N} - 1} \right\} = \frac{1}{{n + 1}}({2^{n + 1}} - 1)$

$\frac{{n!}}{{1!(n - 1)!}} + \frac{1}{{3!}}.\frac{{n!}}{{(n - 3)\,!}} + \frac{1}{{5!}}.\frac{{n!}}{{(n - 5)!}} + ....$

$ = {\,^n}{C_1} + {\,^n}{C_3} + {\,^n}{C_5} + .... = {2^{n - 1}}$.

Thus $\frac{1}{{1!(n - 1)!}} + \frac{1}{{3!(n - 3)!}} + \frac{1}{{5!(n - 5)!}} + ....$$ = \frac{1}{{n!}}{2^{n - 1}}$.

${(x - 2y + 3z)^n}$is ${(1 - 2 + 3)^2} = {2^n}$

 ${2^n} = 128 \Rightarrow n = 7$

Therefore, greatest coefficient in the expansion of ${(1 + x)^7}$is $^7{C_3}$or

$^7{C_4}$because both are equal to 35.

$ \Rightarrow \,\,\,2{(^{15}}{C_8} + {\,^{15}}{C_9} + ... + {\,^{15}}{C_{15}}) = {2^{15}}$ $(​\because \,{\,^n}{C_r} = {\,^n}{C_{n - r}})$ 

==> $^{15}{C_8} + {\,^{15}}{C_9} + ... + {\,^{15}}{C_{15}} = {2^{14}}$

${\left( {1 + \frac{1}{x}} \right)^n} = {C_0} + {C_1}.\frac{1}{x} + {C_2}.\frac{1}{{{x^2}}} + ... + {C_n}\left( {\frac{1}{{{x^n}}}} \right)$

on multiplying both expansions, we get

$\frac{{{{(1 + x)}^{2n}}}}{{{x^n}}} = \sum {C_0^2 + x\sum {{C_0}{C_1} + {x^2}\sum {{C_0}{C_2} + ....} } } $$ + {x^r}\sum {{C_0}{C_r} + .....} $

The various sigma are the coefficient of ${x^0},x,{x^2},.....,{x^r}$ in $L.H.S.$ $\frac{{{{(1 + x)}^{2n}}}}{{{x^n}}}$ or coefficient of ${x^n},{x^{n + 1}},{x^{n + 2}},.....,{x^{n + r}}$ in the expansion of ${(1 + x)^{2n}}$ which occur in ${T_{n + 1,}}{T_{n + 2}},....$ and are

$^{2n}{C_n}{,^{2n}}{C_{n + 1}}{,^{2n}}{C_{n + 2}}{....^{2n}}{C_{n + r}}$etc.

$^{\,2n}{C_{n + 2}} = \frac{{(2\,n\,)\,!}}{{(n - 2)\,!\,(n + 2)\,!}}$

${(1 - x)^n} = {C_0} - {C_1}x + {C_2}{x^2} - {C_3}{x^3} + ..... + {( - 1)^n}{C_n}{x^n}$

$[{(1 + x)^n} - {(1 - x)^n}] = 2\,[{C_1}x + {C_3}{x^3} + {C_5}{x^5} + ...]$

$\frac{1}{2}[{(1 + x)^n} - {(1 - x)^n}] = {C_1}x + {C_3}{x^3} + {C_5}{x^5} + .......$

Put $x = 2$, $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + .....\, = \frac{{{3^n} - {{( - 1)}^n}}}{2}$

Put $n = 2$, $y = {2^2}{\left( {\frac{3}{2}} \right)^4} = 4\,.\,\frac{{81}}{{8 \times 2}} = \frac{{81}}{4}\tilde - 20$

Option  $(a) = {\left( {\frac{{n + 1}}{2}} \right)^3} = \frac{{27}}{8} < y$

Option $(b)  = {\left( {\frac{{n + 1}}{2}} \right)^3} = \frac{{27}}{8} < y$

Option $(c)  = {(2!)^3} = 8 < y$

=$\left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + ... + \left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)$

= $1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}}$
${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = {\left( {\frac{n}{{n + 1}}} \right)^2}$.

We know that, ${(1 + x)^n} = {\,^n}{C_0} + {\,^n}{C_1}{x^1} + {\,^n}{C_2}{x^2} + .... + {\,^n}{C_n}.{x^n}$

Put $x = 1$; ${2^n} = {\,^n}{C_0} + {\,^n}{C_1} + {\,^n}{C_2} + ..... + {\,^n}{C_n}$

Put $n = 20$; ${2^{20}} = {\,^{20}}{C_0} + {\,^{20}}{C_1} + {\,^{20}}{C_2} + ...... + {\,^{20}}{C_{20}}$

${2^{20}} + \,{\,^{20}}{C_{10}} = 2\,[{\,^{20}}{C_0} + {\,^{20}}{C_1} + ...... + {\,^{20}}{C_{10}}]$

${[^{20}}{C_0} + {\,^{20}}{C_1} + ...... + {\,^{20}}{C_{10}}] = {2^{19}} + \frac{1}{2}{\,^{20}}{C_{10}}$

$\sum\limits_{k = 0}^{10} {^{20}{C_k}} = {2^{19}} + \frac{1}{2}{\,^{20}}{C_{10}}$.

Put $x = 2$

==> ${3^n} = {}^n{C_0} + 2.{}^n{C_1} + {2^2}.{}^n{C_2} + {2^3}.{}^n{C_3} + .... + {2^n}{.^n}{C_n}$.

$ = ({x^n} + {\,^n}{C_2}{x^{n - 2}}{a^2} + .......$+${(^n}{C_1}{x^{n - 1}}a + {\,^n}{C_3}{x^{n - 3}}{a^3} + .....)$

=$P + Q$

$\therefore $ ${(x - a)^n} = P - Q,$ As the terms are alter.

$\therefore $ ${P^2} - {Q^2} = (P + Q)(P - Q) = {(x + a)^n}{(x - a)^n}$

${P^2} - {Q^2} = {({x^2} - {a^2})^n}$

==> ${(1 - 3x + 10{x^2})^n} = {(2)^{3n}}$, [Putting $x = 1$] .

Now, $b =$ sum of the coefficients in the expansion of ${(1 + {x^2})^n} = {(1 + 1)^n} = {2^n}$. Clearly, $a = {b^3}$

$\sum\limits_{k = 1}^n k {(n - k + 1)^2} = \sum\limits_{k = 1}^n {k[{{(n + 1)}^2} - 2k(n + 1) + {k^2}]} $

$ = {(n + 1)^2}\sum\limits_{k = 1}^n {k - 2(n + 1)\sum\limits_{k = 1}^n {{k^2} + \sum\limits_{k = 1}^n {{k^3}} } } $

$ = {(n + 1)^2}.\frac{{n(n + 1)}}{2} - 2(n + 1).\frac{{n(n + 1)(2n + 1)}}{6}$$ + \frac{{{n^2}{{(n + 1)}^2}}}{4}$

$ = \frac{{n{{(n + 1)}^2}}}{{12}}[6(n + 1) - 4(2n + 1) + 3n$]

$ = \frac{{n{{(n + 1)}^2}}}{{12}}.(n + 2) = \frac{{n(n + 2){{(n + 1)}^2}}}{{12}}$

Trick : Check by taking $n = 1, 2.$

==> $\sin n\theta = {b_0}{\sin ^0}\theta + {b_1}\sin {\,^1}\theta $

$ + {b_2}{\sin ^2}\theta + {b_3}{\sin ^3}\theta + ..... + {b_n}{\sin ^n}\theta $

==> $\sin n\theta = {b_0} + {b_1}\sin \theta + {b_2}{\sin ^2}\theta + .... + {b_n}{\sin ^n}\theta $

($n$ is an odd integer)

$ = {\,^n}{C_1}\sin \theta .{(1 - {\sin ^2}\theta )^{(n - 1)/2}}$

$ - {\,^n}{C_3}{\sin ^3}\theta {(1 - {\sin ^2}\theta )^{(n - 3)/2}} + ....$

$\therefore \,\,\,{b_0} = 0,{b_1} = $ coefficient of $\sin \theta = {\,^n}{C_1} = n$

($ n -1= n -3$ are all even integers)